Geometry Problem 1530. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

More Details

To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.

To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.

ABEF is a circle with AF as diameter.

ReplyDeleteSo < BFE = < BAE = 45 and so

< AFB = 90 - 45 - 20 = 25

Therefore < AEB = < AFB = 25

Sumith Peiris

Moratuwa

Sri Lanka

25°

ReplyDeleteABEF is a cyclic quadrilateral

(20+45)+(90+AEB)=180

AEB=25

x=25° : https://photos.app.goo.gl/d7LNrJchKUm4RTG66

ReplyDeleteABEF is a cyclic quadrilateral so ^(BEA)=^(BFA)

ReplyDeleteABCD is a square => ^(BCA)=^(ECF)=^(EFC)= 45

so ^(BFA)=^(BEA)=90-45-20=25

Angle ABF = Angle AEF = 90 degrees, hence Quadrilateral ABEF is cyclic with AF as diameter.

ReplyDeleteAngle BAE= Angle BFE =45 deg.

Angle AFB= 90 - 45 - 20 = 25 deg.

Since ABEF is cyclic Angle AEB = Angle AFB = 25 degrees.

25

ReplyDelete